Matching two array elements by rounding up and down

Question

Cemre 2023-4-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1949133-matching-two-array-elements-by-rounding-up-and-down

评论： Cemre 2023-4-19

Dear mathworks community,

I need to create a conditional array that I find quite challenging. I uploaded an example data list and result that shows what I need to have at the end as both Excel file and txt file.

I have a reference list that is a column vector. It contains numbers from -10 to 10 with step size of 1 as in the uploaded file and I have a secondary list that contains numbers from -10 to 10 again, but the step sizes differ and list B can be shorter than list A. What I need to do is this:

I need to match the elements of two lists by rounding up or down the elements of list B and the below condition should apply:

if |listA(n) – listB(m)| < |listA(n) – listB(m+1)|

listA(n) matches with listB(m).

So each member of listA matches with the closest element of listB.

Also, if there is no closer value of listB(m) to listA(n), then, listB(m) can be used multiple times.

For example,

listA(n) = -5, listA(n+1) = -6 ,

listB(m) = -4.6, listB(m+1) = -5.2, listB(m+2) = -5.8 and listB(m+3) = -6.5

listA(n) - listB(m) = |-5-(-4.6)| = 0.4

listA(n) - listB(m+1) = |-5 – (-5.2)| = 0.2

since 0.2 < 0.4 listA(n) matches with listB(m+1)

and listA(n+1) - listB(m+2) = |-6 – (-5.8)| = 0.2,

listA(n+1) - listB(m+3) = |-6-(-6.5)| = 0.5

so listA(n+1) matches with listB(m+2)

Note: The difference between the elements of list A and list B can be greater than 1, still the same rule applies.

Note2: This is not a homework. I just need to calculate rms of the difference between reference list and secondary list.

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Answer 1

dpb 2023-4-18

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1949133-matching-two-array-elements-by-rounding-up-and-down#answer_1218148

移动：dpb 2023-4-18

在 MATLAB Online 中打开

fn='https://www.mathworks.com/matlabcentral/answers/uploaded_files/1359918/data_list.txt';
tT=readtable(fn);
ix=isfinite(tT.B);
tT.C=interp1(tT.B(ix),tT.B(ix),tT.A,'nearest','extrap')
tT = 21×3 table
ABC_____________

    -10     -9.4     -9.4
     -9    -7.85     -9.4
     -8     -5.6    -7.85
     -7     -4.1    -7.85
     -6     -2.6     -5.6
     -5     -1.1     -5.6
     -4      0.4     -4.1
     -3      1.9     -2.6
     -2      3.4     -2.6
     -1      4.9     -1.1
      0      6.4      0.4
      1      7.9      0.4
      2      9.4      1.9
      3      NaN      3.4
      4      NaN      3.4
      5      NaN      4.9